[References] The specific rotation, [a]p, for sucrose is +67. What is the observed rotation for a solution of 0.50 g of sucrose in 10 mL of water in a sample tube having a pathlength of 10 cm? degrees. The observed rotation of a solution of 1.2 g of a compound in 10 mL of water is +1.3 degrees. If the pathlength is 10 cm, what is the specific rotation of the compound?
[References] The specific rotation, [a]p, for sucrose is +67. What is the observed rotation for a solution of 0.50 g of sucrose in 10 mL of water in a sample tube having a pathlength of 10 cm? degrees. The observed rotation of a solution of 1.2 g of a compound in 10 mL of water is +1.3 degrees. If the pathlength is 10 cm, what is the specific rotation of the compound?
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Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**[References]**
The specific rotation, \([α]_D\), for sucrose is \( +67 \). What is the observed rotation for a solution of \( 0.50 \) g of sucrose in \( 10 \) mL of water in a sample tube having a path length of \( 10 \) cm?
\[\_\_\_\_\_\] degrees.
The observed rotation of a solution of \( 1.2 \) g of a compound in \( 10 \) mL of water is \( +1.3 \) degrees. If the path length is \( 10 \) cm, what is the specific rotation of the compound?
\[\_\_\_\_\_\].
---
In this image, there are two questions related to the concept of optical rotation and specific rotation in chemistry. Here’s a detailed explanation for each part of the questions.
### Optical Rotation and Specific Rotation
Optical rotation refers to the rotation of the plane of polarized light by a chiral compound. The degree of this rotation is dependent on the concentration of the compound in solution, the path length through which the light passes, and the specific rotation (\([α]_D\)) of the compound.
#### Formula:
\[ [α]_D = \frac{α}{c \cdot l} \]
Where:
- \([α]_D\) = specific rotation
- \(α\) = observed rotation (in degrees)
- \(c\) = concentration of the solution (in g/mL)
- \(l\) = path length (in dm)
### Explanation of Questions
1. **First Question**:
- Given:
- \([α]_D\) for sucrose = \( +67 \)
- 0.50 g of sucrose in 10 mL of water
- Path length = 10 cm (or 1 dm)
- To find: Observed rotation for this solution.
- Calculation:
- Convert the concentration from g/10 mL to g/mL: \( \frac{0.50 \text{ g}}{10 \text{ mL}} = 0.05 \text{ g/mL} \)
- Use the formula \( α = [α]_D \cdot c \cdot l \)
- Substitute the values: \( α = 67 \times 0.05 \times 1 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff71695ef-d0f4-4003-9f93-ae5ca95ceb18%2Ff21bf6bb-f584-45ea-bbae-df729b5ce903%2F0jp8v3d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**[References]**
The specific rotation, \([α]_D\), for sucrose is \( +67 \). What is the observed rotation for a solution of \( 0.50 \) g of sucrose in \( 10 \) mL of water in a sample tube having a path length of \( 10 \) cm?
\[\_\_\_\_\_\] degrees.
The observed rotation of a solution of \( 1.2 \) g of a compound in \( 10 \) mL of water is \( +1.3 \) degrees. If the path length is \( 10 \) cm, what is the specific rotation of the compound?
\[\_\_\_\_\_\].
---
In this image, there are two questions related to the concept of optical rotation and specific rotation in chemistry. Here’s a detailed explanation for each part of the questions.
### Optical Rotation and Specific Rotation
Optical rotation refers to the rotation of the plane of polarized light by a chiral compound. The degree of this rotation is dependent on the concentration of the compound in solution, the path length through which the light passes, and the specific rotation (\([α]_D\)) of the compound.
#### Formula:
\[ [α]_D = \frac{α}{c \cdot l} \]
Where:
- \([α]_D\) = specific rotation
- \(α\) = observed rotation (in degrees)
- \(c\) = concentration of the solution (in g/mL)
- \(l\) = path length (in dm)
### Explanation of Questions
1. **First Question**:
- Given:
- \([α]_D\) for sucrose = \( +67 \)
- 0.50 g of sucrose in 10 mL of water
- Path length = 10 cm (or 1 dm)
- To find: Observed rotation for this solution.
- Calculation:
- Convert the concentration from g/10 mL to g/mL: \( \frac{0.50 \text{ g}}{10 \text{ mL}} = 0.05 \text{ g/mL} \)
- Use the formula \( α = [α]_D \cdot c \cdot l \)
- Substitute the values: \( α = 67 \times 0.05 \times 1 =
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